actually this can be solved in less time than we would think it will take..

A - starting point where cyclist and motorist meet (9 AM)

B - meeting point after A, this happens when motorist returns (Noon)

C - Pedro - Motorist reaches here at 11.30 (1 hour of waiting).

We need to find time taken by cyclist to reach C from B.

Time taken for cyclist to reach B from A - 180 minutes (9 AM to Noon)

Time taken for Motorist to reach B from C - 30 mins (11.30 to Noon).

We know time taken by Motorist to reach C from A is 90 mins (9 to 10.30).

Hence time taken by motorist from A to B is 60 mins (AC-CB = 90-30 = 60)

so if it takes motorist 60 mins and 180 mins for cyclist for A to B, ratio of speed is 1:3,

Since motorist took 30 mins between C and B, it will take cyclist 90 mins (3 times). Hence answer is 90 mins after noon to reach C or 1.30 PM is the answer.

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santorasantu wrote:

A motor car travelling to point pedro overtakes a cyclist at 9:00A.M. The car reaches point pedro at 10:30 A.M. and after waiting for 1hour, returns meeting the cyclist at noon. When will the cyclist reach point pedro?

A. 1:15PM

B. 1:30PM

C. 1:45PM

D. 2:00PM

E: 1:28PM

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